Solution of Trigonometric Equations

Subject: Optional Mathematics

Overview

A method for finding angles.

(i) First of all, we determine the quadrant where the angle falls. For this, we use the all sin, tan, cos rule.

If sinθ is positive, the angleθ falls in the 1st and 2nd quadrants and if sinθ is negative, the angleθ falls in the 3rd and 4th quadrants. If cosθ is positive,θ lies in the 1st and 4th quadrants and if cosθ is negative,θ lies in the 2nd and 3rd quadrants. if tanθ is positive,θ lies in the first and third quadrants and if tanθ is negative,θ lies in second and fourth quadrants.

(ii) To find the angle in the first quadrant, we find the acute angle which satisfies the equation.

For example, if 2cosθ = 1

then cosθ = $\frac{1}{2}$

or, cosθ = cos600

So, θ = 600

(iii) To find the angle in the second quadrant, we subtract acute angleθ from 1800.

(iv) To find the angle in the third quadrant, we add acute angleθ to 1800

(v) To find the angle in the 4th quadrant, we subtract acute angleθ to 3600

(vi) To find the value ofθ from the equations like sinθ = 0, cosθ = 0, tanθ= 0, sinθ = 1, cosθ = 1, sinθ = -1, cosθ = -1. We should note the following results:

 If sinθ = 0, thenθ = 00, 1800 or 3600 If sinθ = 1, thenθ = 900 If tanθ = 0, thenθ = 00,1800 or 3600 If sinθ = -1, thenθ = 2700 If cosθ = 0, thenθ = 900 or 2700 If cosθ = 1, thenθ = 00 or 3600

If cosθ = -1, then θ = 1800.

Solution of Trigonometric Equations

Equalities like 1 - sin2θ = cos2θ , 1 + cos2θ = 2 cos2θ etc are satisfied by all the values of the angle θ . So, they are called identities.

Let the equalities like 2cosθ = 1,$\sqrt 2$ sinθ = 1 etc. are satisfied by only some values of the angle θ.

A trigonometric ratio of a certain angle has one and only value. But, if the value of a trigonometric ratio is given, the angle is not unique. For example, let us take the equation

2cosx = $\sqrt 3$ or cos x = $\frac{\sqrt 3}{2}$

We know that cos300 = $\frac{\sqrt 3}{2}$

So, x = 300

Again, cos3300= cos(3600 - 300) = cos300 = $\frac{\sqrt 3}{2}$

So, x = 3300

∴ x may be 300 or 3300.

Again if we add 3600 or multiple of 3600 to the angle 300 or 3300, the cosine of any one of these angles will also be $\frac{\sqrt 3}{2}$. So there might be many values of x which satisfy the equation. But we will try to find those angles which lie between 00 and 3600.

A method for finding angles:

(i) First of all, we determine the quadrant where the angle falls. For this, we use the all sin, tan, cos rule.

If sinθ is positive, the angle θ falls in the 1st and 2nd quadrants and if sinθ is negative, the angle θ falls in the 3rd and 4th quadrants. If cosθ is positive, θ lies in the 1st and 4th quadrants and if cosθ is negative, θ lies in the 2nd and 3rd quadrants. If tanθ is positive, θ lies in the first and third quadrants and if tanθ is negative, θ lies in second and fourth quadrants.

(ii) To find the angle in the first quadrant, we find the acute angle which satisfies the equation.

For example, if 2cosθ = 1

then cosθ = $\frac{1}{2}$

or, cosθ = cos600

So, θ = 600

(iii) To find the angle in the second quadrant, we subtract acute angle θ from 1800.

(iv) To find the angle in the third quadrant, we add acute angle θ to 1800

(v) To find the angle in the fourth quadrant, we subtract acute angle θ to 3600

(vi) To find the value of θ from the equations like sinθ = 0, cosθ = 0, tanθ= 0, sinθ = 1, cosθ = 1, sinθ = -1, cosθ = -1. We should note the following results:

 If sinθ = 0, thenθ = 00, 1800 or 3600 If sinθ = 1, thenθ = 900 If tanθ = 0, thenθ = 00,1800 or 3600 If sinθ = -1, thenθ = 2700 If cosθ = 0, thenθ = 900 or 2700 If cosθ = 1, thenθ = 00 or 3600 If cosθ = -1, then θ = 1800.

Things to remember

A method for finding angles

(i) First of all, we determine the quadrant where the angle falls. For this, we use the all sin, tan, cos rule.

(ii) To find the angle in the first quadrant, we find the acute angle which satisfies the equation.

(iii) To find the angle in the second quadrant, we subtract acute angle θ from 1800.

(iv) To find the angle in the third quadrant, we add acute angle θ to 1800

(v) To find the angle in the fourth quadrant, we subtract acute angle θ to 3600

• It includes every relationship which established among the people.
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• common interests and common objectives are not necessary for society.
Using the Quadrant Rule to solve trig. equations : ExamSolutions Maths Tutorials

Here,

2 cos$\theta$ = 1

or, cos$\theta$ = $\frac 12$

or, cos$\theta$ = cos 60° , cos (360° - 60°)

∴ $\theta$ = 60° , 300° Ans

Here,

3 tan$\theta$ = $\sqrt 3$

or, tan$\theta$ = $\frac {\sqrt 3}3$

or, tan$\theta$ = $\frac {\sqrt 3}{\sqrt 3 × \sqrt 3}$

or, tan$\theta$ = $\frac 1{\sqrt 3}$

or, tan$\theta$ = tan 30° , tan (180° + 30°)

∴ $\theta$ = 30° , 210° Ans

Here,

$\sqrt 2$ sec$\theta$ + 2 = 0

or, $\sqrt 2$ sec$\theta$ = - 2

or. sec$\theta$ = $\frac {-2}{\sqrt 2}$

or, sec$\theta$ = -$\frac {\sqrt 2 × \sqrt 2}{\sqrt 2}$

or, sec$\theta$ = -$\sqrt 2$

or, sec $\theta$ = sec (180° - 45°), sec (180v + 45°)

∴ $\theta$ = 135° , 225° Ans

Here,

tanx - sinx = 0

or, $\frac {sinx}{cosx}$ - sinx = 0

or, $\frac {sinx - sinx cosx}{cosx}$ = 0

or, sinx (1 - cosx) = 0

Either,

sinx = 0

sinx = sin 0°

∴ x = 0°

Or,

1 - cosx = 0

or, cosx= 1

or, cosx = cos180°

∴ x = 180°

∴ x = 0° , 180° Ans

Here,

2 sin$\theta$ + 1 = 0

or, 2 sin$\theta$ = - 1

or, sin$\theta$ = -$\frac 12$

or, sin$\theta$ = sin (180° + 30°), sin (360° - 30°)

∴ $\theta$ = 210°, 330° Ans

Here,

cosx + secx = $\frac 52$

or, cosx + $\frac 1{cosx}$ = $\frac 52$

or, $\frac {cos^2x + 1}{cosx}$ = $\frac 52$

or, 2 cos2x + 2 = 5 cosx

or, 2 cos2x - 5 cosx+ 2 = 0

or, 2 cos2x - 4 cosx - cosx + 2 = 0

or, 2 cosx (cosx - 2) - 1 (cosx - 2) = 0

or, (cosx - 2) (2 cosx - 1) = 0

Either,

cosx - 2 = 0

or, cosx = 2 (Impossible)

Or,

2 cosx - 1 = 0

or, 2 cosx = 1

or, cosx = $\frac 12$

or, cosx = cos 60°

∴ x = 60° Ans

Here,

4 sin$\alpha$ = 3 cosec$\alpha$

or, 4 sin$\alpha$ = 3$\frac 1{sin\alpha}$

or, 4 sin2$\alpha$ = 3

or, sin2$\alpha$ = $\frac 34$

∴ sin$\alpha$ =± $\frac {\sqrt 3}2$

Taking positive,

sin$\alpha$ = $\frac {\sqrt 3}2$

or, sin$\alpha$ = sin 60° , sin (180° - 60°)

∴ $\alpha$ = 60° , 120°

Taking negative,

sin$\alpha$ = -$\frac {\sqrt 3}2$

or, sin$\alpha$ = sin (180° + 60°), sin (360° - 60°)

∴ $\alpha$ = 240° , 300°

∴ $\alpha$ = 60°, 120°, 240°, 300° Ans

Here,

cos2$\theta$ - 1 = sin$\theta$

or, 1 - sin2$\theta$ - 1 - sin$\theta$ = 0

or, - sin2$\theta$ - sin$\theta$ = 0

or, - sin$\theta$ (sin$\theta$ - 1) = 0

Either,

sin$\theta$ = 0

or, sin$\theta$ = sin 0°, sin 180°

Or,

sin$\theta$ - 1 = 0

or, sin$\theta$ = 1

or, sin$\theta$ = sin 90°

∴ $\theta$ = 0° , 90° , 180° Ans

Here,

sin$\theta$ - cos$\theta$ = 0

or, sin$\theta$ = cos$\theta$

or, $\frac {sin\theta}{cos\theta}$ = 1

or, tan$\theta$ = tan 45°

∴ $\theta$ = 45° Ans

Here,

cot2x + cosec2x = 3

or, cosec2x - 1 + cosec2x = 3

or, 2 cosec2x = 3 + 1

or, cosec2x = $\frac 42$

or, cosec2x = 2

or, cosec x =± $\sqrt 2$

For positive sign,

cosec x = cosec 45° , cosec (360° - 45°)

For negative sign,

cosec x = cosec (180° - 45°) , cosec (180° + 45°)

∴ x = 45°, 135°, 225° and 315° Ans

Here,

cos 2x = sin x

or, cos 2x = cos (90° - x)

or, 2x = 90° - x

or, 3x = 90°

or, x = $\frac {90°}3$

∴ x = 30° Ans

Here,

sin$\theta$ + cos$\theta$ = $\sqrt 2$

Squaring on both sides:

(sin$\theta$ + cos$\theta$)2= ($\sqrt 2$)2

or, sin2$\theta$ + cos2$\theta$ + 2 sin$\theta$ cos$\theta$ = 2

or, 1 + sin 2$\theta$ = 2

or, sin 2$\theta$ = 2 - 1

or, sin 2$\theta$ = 1

or, sin 2$\theta$ = sin 90°

or, 2$\theta$ = 90°

or, $\theta$ = $\frac {90°}2$

∴ $\theta$ = 45° Ans

Here,

2 cos2$\theta$ = - $\sqrt 3$ cos$\theta$

or, 2 cos2$\theta$ + $\sqrt 3$ cos$\theta$ = 0

or, cos$\theta$ (2 cos$\theta$ + $\sqrt 3$) = 0

Either,

cos$\theta$ = 0

or, cos$\theta$ = cos 90°

Or,

2 cos$\theta$ + $\sqrt 3$ = 0

or, 2 cos$\theta$ = - $\sqrt 3$

or, cos$\theta$ = - $\frac {\sqrt 3}2$

or, cos$\theta$ = cos (180° - 30°)

∴ $\theta$ = 90° and 150° Ans

Here,

cos2$\frac {\theta}2$ - cos$\frac {\theta}2$ + $\frac 14$ = 0

or, (cos$\frac {\theta}2$)2 - 2 . cos$\frac {\theta}2$ . ($\frac 12$)2 = 0

or, (cos$\frac {\theta}2$ - $\frac 12$)2 = 0

or, cos$\frac {\theta}2$ - $\frac 12$ = 0

or, cos$\frac {\theta}2$ = $\frac 12$

or, cos$\frac {\theta}2$ = cos 60°

or, $\frac {\theta}2$ = 60°

or, $\theta$ = 2× 60°

∴ $\theta$ = 120° Ans

Here,

2 cos2$\theta$ = 3 sin$\theta$ = 0

or, 2 (1 - sin2$\theta$) - 3 sin$\theta$ = 0

or, 2 - 2 sin2$\theta$ - 3 sin$\theta$ = 0

or, - 2 sin2$\theta$ - 3 sin$\theta$ + 2 = 0

or, - (2 sin2$\theta$ + 3 sin$\theta$ - 2) = 0

or, 2 sin2$\theta$ + 4 sin$\theta$ - sin$\theta$ - 2 = 0

or, 2 sin$\theta$ (sin$\theta$ + 2) - 1 (sin$\theta$ + 2) = 0

or, (sin$\theta$ + 2) (2 sin$\theta$ - 1) = 0

Either,

sin$\theta$ + 2 = 0

or, sin$\theta$ = - 2 (Impossible)

Or,

2 sin$\theta$ - 1 = 0

or, 2 sin$\theta$ = 1

or, sin$\theta$ = $\frac 12$

or, sin$\theta$ = sin 30° , sin (180° - 30°)

∴ $\theta$ = 30°, 150° Ans

Here,

cos2$\theta$ - sin$\theta$ = $\frac 14$

or, 4 (1 - sin2$\theta$ - sin$\theta$) = 1

or, 4 - 4 sin2$\theta$ - 4 sin$\theta$ - 1 = 0

or, - (4 sin2$\theta$ + 4 sin$\theta$ - 3) = 0

or, 4 sin2$\theta$ + 6 sin$\theta$ - 2 sin$\theta$ - 3 = 0

or, 2 sin$\theta$ (2 sin$\theta$ + 3) - 1 (2 sin$\theta$ + 3) = 0

or, (2 sin$\theta$ + 3) (2 sin$\theta$ - 1) = 0

Either,

2 sin$\theta$ + 3 = 0

or, 2 sin$\theta$ = -3

or, sin$\theta$ = -$\frac 32$

sin$\theta$ is not less than -1. So, it is impossible.

Or,

2 sin$\theta$ - 1 = 0

or, 2 sin$\theta$ = 1

or, sin$\theta$ = $\frac 12$

or, sin$\theta$ = sin 30°, sin (180° - 30°)

∴ $\theta$ = 30° and 150° Ans

Here,

sec$\theta$ . tan$\theta$ = $\sqrt 2$

or, $\frac 1{cos\theta}$ . $\frac {sin\theta}{cos\theta}$ = $\sqrt 2$

or, $\frac {sin\theta}{cos^2\theta}$ = $\sqrt 2$

or, $\frac {sin\theta}{1 - sin^2\theta}$ = $\sqrt 2$

or, sin$\theta$ = $\sqrt 2$ - $\sqrt 2$ sin2$\theta$

or, $\sqrt 2$ sin2$\theta$ + sin$\theta$ - $\sqrt 2$ = 0

or, $\sqrt 2$ sin2$\theta$ + 2 sin$\theta$ - sin$\theta$ - $\sqrt 2$ = 0

or, $\sqrt 2$ sin$\theta$ (sin$\theta$ + $\sqrt 2$) - 1 (sin$\theta$ + $\sqrt 2$) = 0

or, (sin$\theta$ + $\sqrt 2$) ($\sqrt 2$ sin$\theta$ - 1) = 0

Either,

sin$\theta$ + $\sqrt 2$ = 0

or, sin$\theta$ = -$\sqrt 2$ (Impossible)

Or,

$\sqrt 2$ sin$\theta$ - 1 = 0

or, $\sqrt 2$ sin$\theta$ = 1

or, sin$\theta$ = $\frac 1{\sqrt 2}$

or, sin$\theta$ = sin 45° , sin (180° - 45°)

∴ $\theta$ = 45° , 135° Ans

Here,

1 + cos$\theta$ = 2 sin2$\theta$

or, 2 sin2$\theta$ - cos$\theta) - 1 = 0 or, 2 (1 - cos2$$\theta$) - cos$\theta$ - 1 = 0 or, 2 - 2 cos2$\theta$ - cos$\theta$ - 1 = 0 or, - 2 cos2$\theta$ - cos$\theta$ + 1 = 0 or, - (2 cos2$\theta$ + cos$\theta$ - 1) = 0 or, 2 cos2$\theta$ + 2 cos$\theta$ - cos$\theta$ - 1 = 0 or, 2 cos$\theta$ (cos$\theta$ + 1) - 1 (cos$\theta$ + 1) = 0 or, (cos$\theta$ + 1) (2 cos$\theta$ - 1) = 0 Either, cos$\theta$ + 1 = 0 cos$\theta$ = - 1 cos$\theta$ = cos 180° Or, 2 cos$\theta$ - 1 = 0 2 cos$\theta$ = 1 cos$\theta$ = $\frac 12$ cos$\theta$ = cos 60° ∴ $\theta$ = 60° and 180° Ans Here, cos 2$\theta$ = sin$\theta$ or, 1 - 2 sin2$\theta$ = sin$\theta$ or, 2 sin2$\theta$ + sin$\theta$ - 1 = 0 or, 2 sin2$\theta$ + 2 sin$\theta$ - sin$\theta$ - 1 = 0 or, 2 sin$\theta$ (sin$\theta$ + 1) - 1 (sin$\theta$ + 1) = 0 or, (sin$\theta$ + 1) (2 sin$\theta$ - 1) = 0 Either, sin$\theta$ + 1 = 0 sin$\theta$ = - 1 sin$\theta$ = sin 270° Or, 2 sin$\theta$ - 1 = 0 2 sin$\theta$ = 1 sin$\theta$ = $\frac 12$ sin$\theta$ = sin 30°, sin (180° - 30°) ∴ $\theta$ = 30°, 150°, 270° Ans Here, 2 cos2$\theta$ + sin$\theta$ = 2 or, 2 (1 - sin2$\theta$) + sin$\theta$ - 2 = 0 or, 2 - 2 sin2$\theta$ + sin$\theta$ - 2 = 0 or, - 2 sin2$\theta$ + sin$\theta$ = 0 or, - sin$\theta$ (2 sin$\theta$ - 1) = 0 or, sin$\theta$ (2 sin$\theta$ - 1) = 0 Either, - sin$\theta$ = 0 sin$\theta$ = sin 0°, sin 180° Or, 2 sin$\theta$ - 1 = 0 2 sin$\theta$ = 1 sin$\theta$ = $\frac 12$ sin$\theta$ = sin 30°, sin (180° - 30°) ∴ $\theta$ = 0°, 30°, 150° and 180° Ans Here, tan2A - (1 + $\sqrt 3$) tanA + $\sqrt 3$ = 0 or, tan2A - tanA - $\sqrt 3$ tanA + $\sqrt 3$ = 0 or, tanA (tanA - 1) - $\sqrt 3$ (tanA - 1) = 0 or, (tanA - 1) (tanA - $\sqrt 3$) = 0 Either, tanA - 1 = 0 tanA = 1 tanA = tan 45° , tan (180° + 45°) Or, tanA - $\sqrt 3$ = 0 tanA = $\sqrt 3$ tanA = tan 60°, tan (180° + 60°) ∴ A = 45°, 60°, 225°, 240° Ans Here, cos2x = 3 sin2x + 4 cosx or, cos2x - 3 sin2x - 4 cosx = 0 or, cos2x - 3(1 - cos2x) - 4 cosx = 0 or, cos2x - 3 + 3 cos2x - 4 cosx = 0 or, 4 cos2x - 4 cosx- 3 = 0 or, 4 cos2x - 6 cosx + 2 cosx - 3 = 0 or, 2 cosx(2 cosx - 3) + 1 (2 cosx - 3) = 0 or, (2 cos x- 3) (2 cos x + 1) = 0 Either, 2 cos x - 3 = 0 2 cos x = 3 cos x = $\frac 32$ (Impossible) Or, 2 cos x+ 1 = 0 2 cos x = - 1 cos x= -$\frac 12$ cos x = cos (180° - 60°), cos (180° + 60°) ∴ x = 120° , 240° Ans Here, $\sqrt 3$ cotA = $\frac {\sqrt 3}{sinA}$ - 1 or, $\sqrt 3$$\frac {cosA}{sinA}$ = $\frac {\sqrt 3- sinA}{sinA}$ or, $\sqrt 3$cosA sinA = $\sqrt 3$sinA - sin2A or, sin2A + $\sqrt 3$cosA sinA - $\sqrt 3$sinA = 0 or, sinA(sinA + $\sqrt 3$cosA - $\sqrt 3$) = 0 Either, sinA = 0 sinA = sin 0° , sin 180°, sin 360° A = 0°, 180°, 360° (These value does not satisfies in the given equation so we eliminate these values.) or, sinA+ $\sqrt 3$cosA - $\sqrt 3$ = 0 or, sinA + $\sqrt 3$cosA = $\sqrt 3$ or, $\frac {sinA}{\sqrt {1^2 + \sqrt {(3)^2}}}$ + $\frac {\sqrt 3 cosA}{\sqrt {1^2 + \sqrt {(3)^2}}}$ = $\frac {\sqrt 3}{\sqrt {1^2 + \sqrt {(3)^2}}}$ or, $\frac {sinA}{\sqrt {1 + 3}}$ + $\frac {\sqrt 3 cosA}{\sqrt {1 + 3}}$ = $\frac {\sqrt 3}{\sqrt {1 + 3}}$ or, $\frac 12$sinA + $\frac {\sqrt 3}2$cosA = $\frac {\sqrt 3}2$ or, cosA cos30° + sinA sin 30° = $\frac {\sqrt 3}2$ or, cos(A - 30°) = cos 30° , cos(360° - 30°) or, cos (A - 30°) = cos 30°....................(i) or, cos (A - 30°) = cos 330°..................(ii) From (i) A - 30° = 30° A = 30° + 30° A = 60° From (ii) A - 30° = 330° A = 330° + 30° A = 360° Since, A = 360° does not satisfy in the given equation. ∴ A = 60° Ans Here, sin 3x + sin x = 2 sin x or, 2 sin x$\frac {3x + x}2$ cos$\frac {3x - x}2$ = 2 sinx or, 2 sin2x cosx = 2 sinx or, 2× 2 sinx cosxcosx - 2 sinx = 0 or, 2 sinx(2 cos2x - 1) = 0 Either, 2 sinx = 0 sinx = 0 sinx = sin 0° , sin 180° Or, 2 cos2x - 1 = 0 2 cos2x = 1 cos2x = $\frac 12$ cosx=± $\frac 1{\sqrt 2}$ cosx = cos 45° , cos (180° - 45°) ∴ x = 0° , 45° , 135° , 180° Ans Here, $\frac {\sqrt 3}{sin 2\theta}$ + $\frac 1{cos 2\theta}$ = 4 or, $\frac {\sqrt 3 cos 2\theta + sin 2\theta}{sin 2\theta cos 2\theta}$ = 4 or, $\frac {\frac {\sqrt 3}2 cos 2\theta + \frac 12 sin 2\theta}{2 sin 2\theta cos 2\theta}$ = $\frac 44$ or, $\frac {sin 60° cos 2\theta + cos 60° sin 2\theta}{sin 4\theta}$ = 1 or, sin (60° + 2$\theta$) = sin 4$\theta$ or, 60° + 2$\theta$ = 4$\theta$ or, 4$\theta$ - 2$\theta$ = 60° or, 2$\theta$ = 60° or, $\theta$ = $\frac {60°}2$ ∴ $\theta$ = 30° Ans Here, sin 2$\theta$ + sin 4$\theta$ = cos$\theta$ + cos 3$\theta$ or, 2 sin$\frac {2\theta + 4\theta}2$ cos$\frac {2\theta - 4\theta}2$ = 2 cos$\frac {\theta + 3\theta}2$ cos$\frac {\theta - 3\theta}2$ or, 2 sin 3$\theta$ . cos$\theta$ = 2 cos 2$\theta$ . cos$\theta$ or, 2 sin 3$\theta$ . cos$\theta$ - 2 cos 2$\theta$ . cos$\theta$ = 0 or, 2 cos$\theta$ (sin 3$\theta$ - cos 2$\theta$) = 0 Either, 2 cos$\theta$ = 0 cos$\theta$ = 0 cos$\theta$ = cos 90° Or, sin 3$\theta$ - cos 2$\theta$ = 0 sin 3$\theta$ = cos$\theta$ sin 3$\theta$ = sin(90° - 2\theta) 3$\theta$ = 90° - 2$\theta$ 3$\theta$ + 2$\theta$ = 90° 5$\theta$ = 90° $\theta$ = $\frac {90°}5$ $\theta$ = 18° ∴ $\theta$ = 18° , 90° Ans Here, tan$\theta$ + tan 2$\theta$ + $\sqrt 3$ tan$\theta$ tan 2$\theta$ = $\sqrt 3$ or, tan$\theta$ + tan 2$\theta$ = $\sqrt 3$ -$\sqrt 3$ tan$\theta$ tan 2$\theta$ or, tan$\theta$ + tan 2$\theta$ = $\sqrt 3$ (1 - tan$\theta$ tan 2$\theta$) or, $\frac {tan\theta + tan 2\theta}{1 - tan\theta tan 2\theta}$ = $\sqrt 3$ or, tan ($\theta$ + 2$\theta$) = $\sqrt 3$ or, tan 3$\theta$ = $\sqrt 3$ or, tan 3$\theta$ = tan 60°, tan (180° + 60°) Either, 3$\theta$ = 60° $\theta$ = $\frac {60°}3$ $\theta$ = 20° Or, 3$\theta$ = 240° $\theta$ = $\frac {240°}3$ $\theta$ = 80° ∴ $\theta$ = 20°, 80° Ans Here, cot x + cot y = 2 or, $\frac {cos x}{sin x}$ + $\frac {cos y}{sin y}$ = 2 or, $\frac {cos x sin y + cos y sin x}{sin x sin y}$ = 2 or, sin x cos y + cos x sin y = 2 sin x cos y or, sin (x + y) = $\frac {\sqrt 3}2$ [$\because$ 2 sin x cos y = $\frac {\sqrt 3}2$] or, sin (x + y) = sin 60°, sin (180° - 60°) ∴ x + y = 60°, 120° Ans Here, $\sqrt 2$ sin\9\theta$$ = tan $\theta$

Squaring on both sides,

($\sqrt 2$ sin$\theta$)2 = (tan$\theta$)2

or, 2 sin2$\theta$ = tan2$\theta$

or, 2 sin2$\theta$ = $\frac {sin^2\theta}{cos^2\theta}$

or, 2 sin2$\theta$ cos2$\theta$ = sin2$\theta$

or, 2 sin2$\theta$ cos2$\theta$ - sin2$\theta$ = 0

or, sin2$\theta$ (2 cos2$\theta$ - 1) = 0

Either,

sin2$\theta$ = 0

sin$\theta$ = 0

sin$\theta$ = sin 0°, sin 180°

Or,

2 cos2$\theta$ - 1 = 0

2 cos2$\theta$ = 1

cos2$\theta$ = $\frac 12$

cos$\theta$ =± $\frac 1{\sqrt 2}$

Taking Positive sign,

cos$\theta$ = cos 45°, cos (360° - 45°)

Taking Negative sign,

cos$\theta$ = cos (180° - 45°), cos (180° + 45°)

∴ $\theta$ = 0°, 45°, 135°, 135°, 225°, 315° Ans

Here,

$\sqrt 3$ sin$\theta$ - $\sqrt 2$ = cos$\theta$

or, $\sqrt 3$ sin$\theta$ - cos\9\theta\) = $\sqrt 2$

or, $\frac {\sqrt 3 sin\theta}{\sqrt {{\sqrt {(3)^2}} + (-1)^2}}$ -$\frac {cos\theta}{\sqrt {{\sqrt {(3)^2}} + (-1)^2}}$ =$\frac {\sqrt 2}{\sqrt {{\sqrt {(3)^2}} + (-1)^2}}$

or, $\frac {\sqrt 3 sin\theta}{\sqrt {3 + 1}}$ -$\frac {cos\theta}{\sqrt {3 + 1}}$ =$\frac {\sqrt 2}{\sqrt {3 + 1}}$

or, $\frac {\sqrt 3}{\sqrt 4}$ sin$\theta$ - $\frac {cos\theta}{\sqrt 4}$ = $\frac {\sqrt 2}{\sqrt 4}$

or, $\frac {\sqrt 3}2$ sin$\theta$ - $\frac 12$ cos$\theta$ = $\frac {\sqrt 2}2$

or, cos 30° sin$\theta$ - sin30° cos$\theta$ = $\frac {\sqrt 2}{\sqrt 2 × \sqrt 2}$

or, sin$\theta$ cos 30° - cos$\theta$ sin 30° = $\frac 1{\sqrt 2}$

or, sin ($\theta$ - 30°) = sin 45°, sin 135°

Either,

$\theta$ - 30° = 45°

$\theta$ = 45° + 30°

$\theta$ = 75°

OR,

$\theta$ - 30° = 135°

$\theta$ = 135° + 30°

$\theta$ = 165°

∴ $\theta$ = 75° , 165° Ans

Here,

sin 3$\theta$ + sin$\theta$ = sin 2$\theta$

or, 2 sin$\frac {3\theta + \theta}2$ cos$\frac {3\theta - \theta}2$ - sin 2$\theta$ = 0

or, 2 sin$\frac {4\theta}2$ cos$\frac {2\theta}2$ - sin 2$\theta$ = 0

or, 2 sin 2$\theta$ cos$\theta$ - sin 2$\theta$ = 0

or, sin 2$\theta$ (2 cos$\theta$ - 1) = 0

Either,

sin 2$\theta$ = 0

2 sin$\theta$ cos$\theta$ = 0

or, sin$\theta$ = 0

or, sin$\theta$ = sin 0°, sin 180°

$\theta$ = 0° , 180°

or, cos$\theta$ = 0

or, cos$\theta$ = cos 90°

$\theta$ = 90°

or, 2 cos$\theta$ - 1 = 0

or, 2 cos$\theta$ = 1

or, cos$\theta$ = $\frac 12$

or, cos$\theta$ = cos 60°

$\theta$ = 60°

∴ $\theta$ = 0°, 60°, 90° and 180° Ans

Here,

$\sqrt 3$ sin$\theta$ + cos$\theta$ = 1

$\sqrt {\sqrt {(3)^2} + (1)^2}$ = $\sqrt {3 + 1}$ = $\sqrt 4$ = 2

Dividing by 2 on both sides of the given equation,

$\frac {\sqrt 3}2$ sin$\theta$ + $\frac 12$ cos$\theta$ = $\frac 12$

or, sin$\theta$ $\frac {\sqrt 3}2$ + cos$\theta$ $\frac 12$ = $\frac 12$

or, sin$\theta$ cos 30° + cos$\theta$ sin 30° = $\frac 12$

or, sin ($\theta$ + 30°) = sin 30°, sin (180° - 30°), sin (360° + 30°)

Either,

$\theta$ + 30° = 30°

$\theta$ = 30° - 30°

$\theta$ = 0°

Or,

$\theta$ + 30° = 150°

$\theta$ = 150° - 30°

$\theta$ = 120°

Or,

$\theta$ + 30° = 390°

$\theta$ = 390° - 30°

$\theta$ = 360°

∴ $\theta$ = 0°, 120°, 360° Ans

Here,

$\sqrt 3$ cos x + sin x

Dividing the equation by$\sqrt {\sqrt {(3)^2} + (1)^2}$ = $\sqrt {3 + 1}$ = $\sqrt 4$ = 2

or, $\frac {\sqrt 3}2$ cos x + $\frac 12$ sin x = $\frac {\sqrt 3}2$

or, sin 60° cos x + cos 60° sin x = $\frac {\sqrt 3}2$

or, sin (60° + x) = sin 60°, sin (180° - 60°), sin (360° + 60°)

Either,

60° + x = 60°

x = 60° - 60°

x = 0°

Or,

60° + x = 120°

x = 120° - 60°

x = 60°

Or,

60° + x = 420°

x = 420° - 60°

x = 360°

∴ x = 0°, 60°, 360° Ans

Here,

cos 3x + cos x = 2 cos x

or, 2 cos$\frac {3x + x}2$ cos$\frac {3x - x}2$ = 2 cos x

or, 2 cos 2x cos x = 2 cos x

or, 2 cos 2x cos x - 2 cos x = 0

or, 2 cos x (cos 2x - 1) = 0

or,cos x (cos 2x - 1) = 0

Either,

cos x = 0

cos x = cos 90°, cos 270°

Or,

cos 2x - 1 = 0

cos 2x = 1

cos 2x = cos 0°, cos 360°

2x = 0° , 360°

x = 0° , 180°

∴ x = 0°, 90°, 180°, 270° Ans

Here,

$\sqrt 3$ sin$\alpha$ - cos$\alpha$ = $\sqrt 2$

or, $\frac {\sqrt 3 sin\alpha}{\sqrt {{\sqrt {(3)^2}} + (-1)^2}}$ -$\frac {cos\alpha}{\sqrt {{\sqrt {(3)^2}} + (-1)^2}}$ =$\frac {\sqrt 2}{\sqrt {{\sqrt {(3)^2}} + (-1)^2}}$

or, $\frac {\sqrt 3 sin\alpha}{\sqrt {3 + 1}}$ -$\frac {cos\alpha}{\sqrt {3 + 1}}$ =$\frac {\sqrt 2}{\sqrt {3 + 1}}$

or, $\frac {\sqrt 3}{\sqrt 4}$ sin$\alpha$ - $\frac {cos\alpha}{\sqrt 4}$ = $\frac {\sqrt 2}{\sqrt 4}$

or, $\frac {\sqrt 3}2$ sin$\alpha$ - $\frac 12$ cos$\alpha$ = $\frac {\sqrt 2}2$

or, cos 30° sin$\alpha$ - sin30° cos$\alpha$ = $\frac {\sqrt 2}{\sqrt 2 × \sqrt 2}$

or, sin$\alpha$ cos 30° - cos$\alpha$ sin 30° = $\frac 1{\sqrt 2}$

or, sin ($\alpha$ - 30°) = sin 45°, sin 135°

Either,

$\alpha$ - 30° = 45°

$\alpha$ = 45° + 30°

$\alpha$ = 75°

OR,

$\alpha$ - 30° = 135°

$\alpha$ = 135° + 30°

$\alpha$ = 165°

∴ $\alpha$ = 75° , 165° Ans